Do I have to use a maximum principle?

Question

Let $\Omega\subset C^0$ a bounded domian in $\mathbb{R}^2$. Let $u\in C^2(\Omega)\cap C(\overline{\Omega})$ be a non negative classical solution of $$ (1+x^2)u_{xx}-2xu_{xy}+(1+u)u_{yy}-(1+u^2)u_x+(1+u_x)u_y-u=1\text{ in }\Omega,\\u(x,y)=\frac{\sin^2(x)}{1+y^2}\text{ on }\partial\Omega. $$ Show that $(0\leq) y\leq 1$ for $(x,y)\in\Omega$.

Hello! Could you please give me a hint how to show that? In the reading we had a maximum principle for linear elliptic PDE of degree 2, but this PDE here is not linear.

Which maximum principle do I have to apply?

Answer 1

Whether or not you have a ready-made maximum principle that applies to this PDE, you can argue from the fundamentals. The maximum of $u$ is attained somewhere on $\overline{\Omega}$. Suppose $\max u>1$, then it's attained at some interior point $(a,b)\in \Omega$. (The boundary values are bounded by $1$.) At the point $(a,b)$ the PDE yields $$(1+a^2)u_{xx}-2a u_{xy}+(1+u)u_{yy}=1+u>0 \tag{1}$$

This contradicts the second derivative test; the explanation is hidden below.

The second derivative test says that the Hessian matrix $H=\begin{pmatrix} u_{xx} & u_{xy} \\ u_{xy} & u_{yy} \end{pmatrix}$ is negative semidefinite at $(a,b)$. The left side of (1) is the sum of $ (a,-1) H (a,-1)^T $, $u_{xx}$, and $uu_{yy}$: all these quantities must be nonpositive.